On using elementary row operation ${R_1} \to {R_1} - 3{R_2}$ in the following matrix equation $\left[ {\begin{array}{llllllllllllllllllll}4&2\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&2\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$, we have

We have, $\left[ {\begin{array}{llllllllllllllllllll}4&2\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&2\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$

Using elementary row operation ${R_1} \to {R_1} - 3{R_2}$,

$\left[ {\begin{array}{cccccccccccccccccccc}{ - 5}&{ - 7}\\3&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 7}\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$

Since, on using elementary row operation on $X = AB$,

we apply these operation simultaneously on $X$ and on the first matrix $$A$$ of the product $AB$ on RHS.